From my book:
6. With regard to the limit $\displaystyle \lim_{x\rightarrow 5}x^2=25$,
(a) Show that $\left |x^2-25 \right |<11\left |x-5 \right |$ if $4<x<6$
(b) Find a $\delta$ such that $\left |x^2-25 \right |<10^{-3}$ if $0<\left |x-5 \right |<\delta$
(c) Give a rigorous proof of the limit by showing that $\left |x^2-25 \right |<\epsilon$ if $0<\left|x-5\right|<\delta$, where $\delta$ is the smaller of $\frac{\epsilon}{11}$ and $1$.
For (a):
$|x^2-25|<11|x-5|$
$|x+5||x-5|<11|x-5|$
$|x+5|<11$
$-16<x<6$
And ignoring the interval $[-16,4)$, $4<x<6$
For (b) and (c) I am unsure.
Hint: let's find a $\delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < \delta'$. If we select $\delta = \min\{\delta',1\}$, then we can guarantee that if $|x-5| < \delta$, then (since $x$ is necessarily between $4$ and $6$), $$ |x^2 - 5| < 11|x - 5| < 10^{-3} $$ For (c), all you should need to do is substitute $\epsilon$ for $10^{-3}$.